Perform n steps to convert every digit of a number in the format [count][digit]

Given a number num as a string and a number N. The task is to write a program which converts the given number num to another number after performing N steps. At each step, every digit of num will be written in the format [count][digit] in the new number, where count is the number of times a digit occurs consecutively in num.

Examples:

Input: num = “123”; n = 3
Output: 1321123113
For, n = 1: 123 becomes 1 time 1, 1 time 2, 1 time 3, hence number 111213
For, n = 2: 3 times 1, 1 time 2, 1 time 1, 1 time 3, hence number 31121113
For, n = 3: 1 time 3, 2 times 1, 1 time 2, 3 times 1, 1 time 3, hence number 1321123113

Input: num = “1213”; n = 1
Output: 11121113

Approach: Parse the string’s characters as a single digit and maintain a count for that digit till a different digit is found. Once a different digit is found, add the count of the digit to the new string and number to it. Once the string is parsed completely, recur for the function again with this new string till n steps are done.

Below is the implementation of the above approach:

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// C++ program to convert number
// to the format [count][digit] at every step
#include <bits/stdc++.h>
using namespace std;
  
// Function to perform every step
void countDigits(string st, int n)
{
  
    // perform N steps
    if (n > 0) {
        int cnt = 1, i;
        string st2 = "";
  
        // Traverse in the string
        for (i = 1; i < st.length(); i++) {
            if (st[i] == st[i - 1])
                cnt++;
            else {
                st2 += ('0' + cnt);
                st2 += st[i - 1];
                cnt = 1;
            }
        }
  
        // for last digit
        st2 += ('0' + cnt);
        st2 += st[i - 1];
  
        // recur for current string
        countDigits(st2, --n);
    }
  
    else
        cout << st;
}
  
// Driver Code
int main()
{
  
    string num = "123";
    int n = 3;
  
    countDigits(num, n);
  
    return 0;
}

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Output:

1321123113


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